\newproblem{lay:1_3_27}{
   % Problem identification
	 \begin{large}
 	   \hspace{\fill}\newline
     \textbf{Lay, 1.3.27}
	 \end{large}
	 \\
   \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

   % Problem statement
   A mining company has two mines. One's day operation at mine \#1 produces ore that contains 30 metric tons of copper and 600 kg of silver, while 
	 one day's operation at mine \#2 produces ore that containes 4 metric tones of copper and 380 kg of silver. Let $\mathbf{v}_1=\begin{pmatrix}30\\600\end{pmatrix}$
	 and $\mathbf{v}_2=\begin{pmatrix}40\\380\end{pmatrix}$. Then, $\mathbf{v}_1$ and $\mathbf{v}_2$ represent the output per day at mines \#1 and \#2, respectively.
	\begin{enumerate}[a]
		\item What physical interpretation can be given to the vector $5\mathbf{v}_1$?
		\item Suppose the company operates mine \#1 for $x_1$ days, and mine \#2 for $x_2$ days. Write a vector equation whose solution gives the number of days each mine
		      should operate in order to produce 240 tons of copper and 2824 kg of silver.
		\item Solve the previous equation
	\end{enumerate}
}{
   % Solution
	 \begin{enumerate}[a]
		  \item $5\mathbf{v}_1$ is the production of copper and silver of mine \#1 after 5 days of operation.
			\item The vector equation sought is
			      \begin{center}
						   $x_1\mathbf{v}_1+x_2\mathbf{v}_2=\begin{pmatrix}240\\2824\end{pmatrix}$
						\end{center}
						or what is the same
			      \begin{center}
						   $\begin{pmatrix}30x_1+40x_2\\600x_1+380x_2\end{pmatrix}=\begin{pmatrix}240\\2824\end{pmatrix}$
						\end{center}
			\item The solution of this equation is $x_1=1.7270$ and $x_2=4.7048$, as can be easily checked
			      \begin{center}
						   $\begin{pmatrix}30\cdot 1.7270+40\cdot 4.7048\\600\cdot 1.7270+380\cdot 4.7048\end{pmatrix}=\begin{pmatrix}240\\2824\end{pmatrix}$
						\end{center}
	 \end{enumerate}
}

\useproblem{lay:1_3_27}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
